For the reaction:2 BaO₂(s) ⇌ 2BaO(s) + O₂(g);ΔH = +ve. In equilibrium condition, pressure of O₂ is dependent on

mass of BaO₂
mass of BaO
temperature of equilibrium
mass of BaO₂ and BaO both

Solution

For the reaction
2BaO(s) ⇌ 2BaO(s) + O₂(g); ΔH = +ve.
In equilibrium kP=pO₂
Hence, the value of equilibrium constant depends only upon partial pressure of O₂. Further on increasing temperature formation of O₂ increases as this is an endothermic reaction. Hence, pressure of O₂ is dependent on temperature.

Solution

Solution

When volume is increased the conc. decreases & the equilibrium shifts in the direction where more moles are formed.

pH of a solution containing 0.3 M HX and 0.1M X^– (K_b for X^–= 1.0 × 10^-5) is –

5 + log 3
5 – log 3
9– log 3
9 + log 3

Solution

If the Kp for the equilibrium,M.5H₂O(s) ⇌ M.3H₂O(s) + 2H₂O(g) is 1 × 10^-4. Then M.5H₂O(s) will show efflorescence when it is exposed to an atmosphere where vapour pressure of water is –

more than 10^-2 atm
below 10^-2 atm
more than 10^-4 atm
below 10^-4 atm

Solution

Solid AgNO₃ is slowly added to a solution containing each of 0.01 M NaCl and 0.001 M NaBr. What will be the concentration of Cl^– ions in solution when AgBr will just start to precipitate ? K_sp(AgBr) = 3.6 × 10^-13, K_sp (AgCl) = 1.8 × 10^-10.

1.8 × 10^-7
3.6 × 10^-10
0.01
2 × 10^-4

Solution

K_sp(AgBr)< K_sp(AgCl)Therefore, AgBr will precipitate first and at that time all Cl–will be present.

Three sparingly soluble salts A₂B, AB and AB₃ have the same values of solubility products (K_sp). In a saturated solution the correct order of their solubilities is

AB₂ > AB >> A₂B
AB₃ > A₂B > AB
AB > AB₃ > A₂B
AB > A₂B > AB₃

Solution

Solubilities of three sparingly soluble salts XY (K_sp), XY₂(K’_sp) and X₂Y₃(K”_sp) are equal in water. What will be the correct order of their solubility products –

K_sp < K'_sp < K''_sp
K_sp < K''_sp < K'_sp
K''_sp < K'_sp < K_sp
K''_sp < K_sp < K'_sp

Solution

Since for sparingly soluble salts, solubility (s) will be less than one. Therefore as the powers of s increases in K_sp expression, magnitude of K_sp will become smaller.

Solution

In a closed system : A_(s) ⇌ 2B_(g) + 3C_(g), if the partial pressure of C is doubled, then partial pressure of B will be:

two times the original value
one-half of its original value
\(\frac{1}{2\sqrt{2}}\) times the original value
\(2\sqrt{2}\) times its original value

Solution

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